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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 85 483 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 507-o'rinni va Hindiston mintaqasida 3 480-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 85 483 obunachiga ega bo‘ldi.

24 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 369 ga, so‘nggi 24 soatda esa -35 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.60% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.55% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 2 222 marta ko‘riladi; birinchi sutkada odatda 474 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 3 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Kanal uchun tavsif kiritilmagan.

Yuqori yangilanish chastotasi (oxirgi ma’lumot 25 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

85 483
Obunachilar
-3524 soatlar
-2687 kunlar
-1 36930 kunlar
Postlar arxiv
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Amazon Off Campus Hiring 2024 : Recruitment for Freshers | Register Now Eligibility: Testing Associate, Device OS: Bachelor’s degree Knowledge of QA methodology and tools Preferred: B.E,B.Tech. , MSC IT and MCA Good Communication Testing Knowledge Apply:- https://www.allcoding1.com/2024/02/amazon-off-campus-hiring-2024.html Telegram:- @allcoding1_official

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All this 300 rupees only Contact:- @meterials_available
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int solve(int N, vector<int> A, int T) {     int xa = 0;     for(int i=0; i<N; i++) {         xa ^= A[i];     }     int d = xa ^ T;     if(d == 0) return 0;     int ops = 0;     while(d > 0) {         ops += d & 1;         d >>= 1;     }     return ops; } Flip or flop ✅ Telegram:- @meterials_available

All this 300 rupees only Contact:- @meterials_available
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#include <bits/stdc++.h> using namespace std; int getMinimumMoves(string word) {     unordered_map<char, int> freq;     int moves = 0;     for (char c : word) {         freq[c]++;     }     for (auto& entry : freq) {         moves += entry.second / 2;     }     return moves; } Minimum length word ✅

#include <iostream> #include <vector> #include <unordered_map> #include <sstream> using namespace std; vector<string> computeParameterValue(vector<vector<string>>& sources) {     unordered_map<string, string> final_parameters;     vector<string> order_of_keys;     for (auto& source : sources) {         for (auto& pair : source) {             stringstream ss(pair);             string key, value;             getline(ss, key, ':');             getline(ss, value, ':');             if (final_parameters.find(key) == final_parameters.end()) {                 order_of_keys.push_back(key);             }             final_parameters[key] = value;         }     }     vector<string> final_parameters_list;     for (auto& key : order_of_keys) {         final_parameters_list.push_back(final_parameters[key]);     }     return final_parameters_list; } Stock Analysis ✅

static int getAnagramPeriod(String input_str) {         int n = input_str.length();                 for (int period = 1; period <= n; period++) {             if (n % period == 0) {                 String subString = input_str.substring(0, period);                 boolean valid = true;                 for (int i = 0; i < n; i += period) {                     if (!isAnagram(subString, input_str.substring(i, i + period))) {                         valid = false;                         break;                     }                 }                 if (valid) {                     return period;                 }             }         }         return -1;     }     static boolean isAnagram(String str1, String str2) {         char[] charArray1 = str1.toCharArray();         char[] charArray2 = str2.toCharArray();         Arrays.sort(charArray1);         Arrays.sort(charArray2);         return Arrays.equals(charArray1, charArray2);     } Anagram Period ✅

Repost from allcoding1
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All this 300 rupees only Contact:-  @meterials_available

All this 300 rupees only Contact:- @meterials_available
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